Inequalities

[hardy wind]

I need help with all these problems because all of them seem really confusing for me.

[compact roost]

what is confusing, kind mr gentleman sir?

what did you try on 63?

[hardy wind]

i genuinely dont know what to do for 63

[fickle pewter]

Since b > a, b can be equal to a+x, where x is a positive real number

The second equation becomes (2a + x)/2

The third side becomes a+x

You can also put everything over a common denominator

a -> 2a / 2
(2a+x)/ 2 stays as is
(a+x) -> 2a + 2x
Multiply two to get
2a < 2a + x < 2a + 2x
(Remember that x is a positive number)

For 65)
|x| = x if x is positive/zero, -x if x is negative
|x| is always positive

If a and b are both positive or both negative
|ab| = ab
|a|•|b| = ab or -a•-b which is ab
If either a or b is negative (one is negative, one is positive)
|ab| = -ab
|a|•|b| = -a•b or a•-b, both equal -ab

Aw dam I forgot about spoiler text

For 67, use the same trick in 63 (remember that a positive • positive = positive, positive + positive = positive, and positive > 0)

For 69 (nice), a rational is any fraction a/b where both a and b are integers.
Remember that an integer times,plus, or minus an integer is ALWAYS an integer

@hardy wind hope this helps

[hardy wind]

it does thanks

[civic token]

is anyone here avaliable to explain why you have to flip the inequality sign when dealing with a negative

[dim path]

Let's imagine we have -a > -b. Now, instead of multiplying by -1, we do this: -a > -b 0 > a - b (add a to both sides) b > a (add b to both sides)

[civic token]

okay say we have -2x < 2
I divide -2 on both sides to get
x < -1 or x > -1
I get confused on which sign

@dim path

[dim path]

So don't divide. Add. Specifically, add 2x - 2.

Or divide by 2 first and add x - 1.